What are the solutions to the equation w^2+8w-48=0 ?

The solutions to the equation w^2+8w-48=0 are w=4,w=-12

Find the zeros of w^2+8w-48=0

The zeros of w^2+8w-48=0 are w=4,w=-12

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w^{2} + 8 w - 48 = 0

w = 4, w = - 12

Solution steps

w^{2} + 8 w - 48 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 48) = 16

w_{1, 2} = \frac{- 8 \pm 16}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 8 + 16}{2 \cdot 1}, w_{2} = \frac{- 8 - 16}{2 \cdot 1}

w = \frac{- 8 + 16}{2 \cdot 1} : 4

w = \frac{- 8 - 16}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

w = 4, w = - 12

r^{2} + 10 r = - 7

2.32 = - 0.0007 h^{2} + 0.015 h + 3.02

t, t^{2} = 2 p + a s

x^{2} - 4 x - 10 = 2

x, 3 x^{2} - 3 x + 9 = 0

What are the solutions to the equation w^2+8w-48=0 ?
The solutions to the equation w^2+8w-48=0 are w=4,w=-12
Find the zeros of w^2+8w-48=0
The zeros of w^2+8w-48=0 are w=4,w=-12